Mean Value Theorem Solution 4

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SOLUTION 4: We are given the function

$f(x) = \displaystyle{ x \over 1+x }$ and the interval

$[1, 3]$ . This function is continuous on the closed interval

$[1, 3]$ since it is the quotient of continuous functions

$y=x$ (polynomial) and

$y=1+x$ (polynomial). The derivative of

$f$ is

$f'(x) = \displaystyle{ (1+x)(1) - x(1) \over (1+x)^2 } = { 1 \over (1+x)^2 }$ We can now see that

$f$ is differentiable on the open interval

$(1, 3)$ . The assumptions of the Mean Value Theorem have now been met. Let's apply the Mean Value Theorem and find all possible values of

$c$ in the open interval

$(1, 3)$ . Then

$f'(c)= \displaystyle{ f(3)-f(1) \over 3-1 } \ \ \ \ \longrightarrow$

$\displaystyle{ 1 \over (1+c)^2 } = \displaystyle{ { (3) \over 1+(3) } - { (1) \over 1+(1) } \over 3-1 } \ \ \ \ \longrightarrow$

$\displaystyle{ 1 \over (1+c)^2 } = \displaystyle{ { 3 \over 4 } - { 1 \over 2 } \over { 2 \over 1 } } \ \ \ \ \longrightarrow$

$\displaystyle{ 1 \over (1+c)^2 } = \displaystyle{ \Big( { 3 \over 4 } - { 2 \over 4 } \Big) \cdot { 1 \over 2 } } \ \ \ \ \longrightarrow$

$\displaystyle{ 1 \over (1+c)^2 } = \displaystyle{ 1 \over 8 } \ \ \ \ \longrightarrow$

$(1+c)^2 = 8 \ \ \ \ \longrightarrow$

$1+c = \pm \sqrt{8} = \pm 2 \ \sqrt{2} \ \ \ \ \longrightarrow$

$c = \pm 2 \ \sqrt{2} -1 \ \ \ \ \longrightarrow$

$c = 2 \ \sqrt{2} -1 \approx 1.828 \ \ \ or \ \ \ -2 \ \sqrt{2} -1 \approx -3.828 \ \ \ (-3.828 \ is \ NOT \ in \ the \ interval \ (1, 3).) \ \ \ \ \longrightarrow$

$c = 2 \ \sqrt{2} -1 \approx 1.828 \ \ \ \ \longrightarrow$

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